If y = fleft( frac{2x - 1}{x^2 + 1} right) an | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given $y = f\left( \frac{2x - 1}{x^2 + 1} \right).$ Let $t = \frac{2x - 1}{x^2 + 1}.$By the chain rule,$\frac{dy}{dx} = f'(t) \cdot \frac{dt}{dx}.

Vedclass · Accepted Answer

$\frac{-2x^2 + 2x + 2}{(x^2 + 1)^2} \sin \left( \frac{2x - 1}{x^2 + 1} \right)^2$

Vedclass · Answer

(D) Given $y = f\left( \frac{2x - 1}{x^2 + 1} \right).$ Let $t = \frac{2x - 1}{x^2 + 1}.$By the chain rule,$\frac{dy}{dx} = f'(t) \cdot \frac{dt}{dx}.$Since $f'(x) = \sin(x^2),$ we have $f'(t) = \sin(t^2) = \sin \left( \frac{2x - 1}{x^2 + 1} \right)^2.$Now,calculate $\frac{dt}{dx}$ using the quotient rule:$\frac{dt}{dx} = \frac{(x^2 + 1)(2) - (2x - 1)(2x)}{(x^2 + 1)^2} = \frac{2x^2 + 2 - 4x^2 + 2x}{(x^2 + 1)^2} = \frac{-2x^2 + 2x + 2}{(x^2 + 1)^2}.$Therefore,$\frac{dy}{dx} = \frac{-2x^2 + 2x + 2}{(x^2 + 1)^2} \sin \left( \frac{2x - 1}{x^2 + 1} \right)^2.$

If $y = f\left( \frac{2x - 1}{x^2 + 1} \right)$ and $f'(x) = \sin(x^2),$ then $\frac{dy}{dx} = $

Similar Questions

$\frac{d}{dx} \log_{\sqrt{x}} \left(\frac{1}{x}\right)$ is equal to

$(1+\Delta)^{n} f(a)$ is equal to

The differential coefficient of ${x^6}$ with respect to ${x^3}$ is

If $y = \cos^{-1}(\tanh x) + \sinh(\sin 6x)$, then $\frac{dy}{dx} =$

Suppose that $h(x) = f(x) \cdot g(x)$ and $F(x) = f(g(x))$,where $f(2) = 3$,$g(2) = 5$,$g'(2) = 4$,$f'(2) = -2$,and $f'(5) = 11$. Then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module